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A350461 Number of ways to choose a subset of size n from [2n] and arrange its elements into a set of lists.

%I A350461 #59 May 14 2022 19:50:40
%S A350461 1,2,18,260,5110,126252,3743124,129156456,5075323110,223484406860,
%T A350461 10889720208796,581327564001912,33721264023340348,2111076358455927800,
%U A350461 141812884019465389800,10171645727323281955920,775654703427461395949190,62649431136582816113115660
%N A350461 Number of ways to choose a subset of size n from [2n] and arrange its elements into a set of lists.
%H A350461 Alois P. Heinz, <a href="/A350461/b350461.txt">Table of n, a(n) for n = 0..361</a>
%F A350461 a(n) = binomial(2*n,n) * A000262(n) = A000984(n) * A000262(n).
%F A350461 a(n) = A129652(2n,n).
%e A350461 a(2) = 18: 12, 21, 1|2, 13, 31, 1|3, 14, 41, 1|4, 23, 32, 2|3, 24, 42, 2|4, 34, 43, 3|4.
%p A350461 b:= proc(n) option remember; `if`(n=0, 1, add(
%p A350461       b(n-j)*binomial(n-1, j-1)*j!, j=1..n))
%p A350461     end:
%p A350461 a:= n-> binomial(2*n, n)*b(n):
%p A350461 seq(a(n), n=0..20);
%t A350461 a[n_] := If[n==0, 1, ((2n)!/n!) Sum[Binomial[n-1, j]/(j+1)!, {j, 0, n-1}]];
%t A350461 Table[a[n], {n, 0, 20}] (* _Jean-François Alcover_, May 14 2022, from 1st formula *)
%Y A350461 Cf. A000262, A000984, A129652.
%K A350461 nonn
%O A350461 0,2
%A A350461 _Alois P. Heinz_, Feb 22 2022